Welcome to Software Development on Codidact!
Will you help us build our independent community of developers helping developers? We're small and trying to grow. We welcome questions about all aspects of software development, from design to code to QA and more. Got questions? Got answers? Got code you'd like someone to review? Please join us.
Comments on When does it not work to dereference the pointer for sizeof during malloc?
Post
When does it not work to dereference the pointer for sizeof during malloc?
Background
This is kind of a subquestion to How to properly use malloc?
When allocating, there are basically two common ways of using the sizeof operator:
int *p;
p = malloc(n * sizeof *p); // Method 1: Dereference the pointer
p = malloc(n * sizeof(int)); // Method 2: Explicitly use the type
I personally prefer method 1, because it reduces code duplication.* And the rule is pretty simple. Take whatever you have on the left of the equal sign, and add an asterisk to get the argument for sizeof. It also works for 2D or 3D:
int ***p;
p = malloc(x * sizeof *p); // Add asterisk to p
for(int i=0; i<x; i++) {
p[i] = malloc(y * sizeof *p[i]); // Add asterisk to p[i]
for(int j=0; j<y; j++) {
p[i][j] = malloc(z * sizeof *p[i][j]); // Add asterisk to p[i][j]
}
}
* In the sense that if you have multiple malloc calls where you assign to the pointer p, then if you want to change the type of the pointer, you would need to find ALL malloc calls and change them.
Array parameters
I know of one instance where it "doesn't work", and that is when you have complex parameters to functions. Like this:
void foo(int n, int32_t p[5][5]) {
But in this case, I'd say that the problem is that when arrays are declared as function arguments like that, they are not arrays. Indeed, the C syntax here is a bit strange. The equivalent declaration int32_t (*p)[5] works as expected with this rule. Because the type of p is pointer to array 5 of int32_t. In other words, I expect sizeof *p to be 20, which it also is. The solution to the above problem is to - explicitly in code or mentally in your head - introduce a temporary variable:
void foo(int n, int32_t p[5][5]) {
int32_t (*tmp)[5];
tmp = malloc(n * sizeof *tmp);
void pointers
I know that it does not work if you have void pointers. For example:
void *p;
p = malloc(n * sizeof *p); // Error: A void pointer cannot be dereferenced
p = malloc(sizeinbytes); // Works fine
However, even if this isn't a very extreme and unusual case, this is indeed a pretty special case. And it's also pretty obvious that it's an exception. A C programmer should know that void pointers cannot be dereferenced. However, do note that both gcc and clang required compiling with -pedantic to warn about this.
Flexible array members
Another case is if you have a pointer to struct where the struct contains a flexible array. But although this indeed is a very valid real world use case where this approach would not work, it's also abundantly obvious that it is an exception, since the whole foundation around flexible array members is that you manually Furthermore, the rule can be shoehorned in if you want to. At least for the case where you want several instances of the struct and the flexible array equally big for each of them. Like this:
struct s {
int x;
char y;
double z;
long a[];
};
// m is how many elements the array s::a should have in each instance of s
// n is how many instances of s we want
struct s *p = malloc(n * (sizeof(*p) + m * sizeof *(p->a)));
Question
So my question is, when does this approach with just adding an asterisk to the pointer not work? I'm interested in both realistic use cases and examples created just to break this rule of thumb. Please specify if your example is a real issue or just a theoretical one.
3 comment threads